Q45 / 120 · JEE Advanced 2006 · Mathematics

If $a, b$ and $c$ are the sides of a $\triangle A B C$ such that $x^2-2(a+b+c) x+3 \lambda(a b+b c+c a)=0$ has real roots, then

  • Correct answer
    A
    $\lambda < \frac{4}{3}$
  • B
    $\lambda>\frac{5}{3}$
  • C
    $\lambda \in\left(\frac{4}{3}, \frac{5}{3}\right)$
  • D
    $\lambda \in\left(\frac{1}{3}, \frac{5}{3}\right)$

Solution

Since, roots are real, $D \geq 0$ $$ \begin{array}{ll} \Rightarrow & 4(a+b+c)^2-12 \lambda(a b+b c+c a) \geq 0 \\ \Rightarrow & (a+b+c)^2 \geq 3 \lambda(a b+b c+c a) \\ \Rightarrow & a^2+b^2+c^2 \geq(a b+b c+c a)(3 \lambda-2) \\ \Rightarrow & 3 \lambda-2 \leq \frac{a^2+b^2+c^2}{a b+b c+c a} \\ \text { where, } a^2+b^2 \geq 2 a b, b^2+c^2 \geq 2 b c \text { and } c^2+a^2 \geq 2 c a \\ \Rightarrow & a^2+b^2+c^2 \geq(a b+b c+c a) \\ \therefore & \frac{a^2+b^2+c^2}{a b+b c+c a} \geq 1 \end{array} $$ Also, $\quad \cos A=\frac{b^2+c^2-a^2}{2 b c} < 1$ $$ \Rightarrow \quad b^2+c^2-a^2 < 2 b c $$ Similarly, $c^2+a^2-b^2 < 2 c a$ and $a^2+b^2-c^2 < 2 a b$ $$ \Rightarrow a^2+b^2+c^2 < 2(a b+b c+c a) \text { or } \frac{a^2+b^2+c^2}{a b+b c+c a} < 2 $$ From Eqs. (i), (ii) and (iii) $$ \begin{array}{lrl} \therefore & 3 \lambda-2 & < 2 \\ \Rightarrow & \lambda < \frac{4}{3} \end{array} $$

Hence, the correct answer is option: 'A'.

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